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The question in the image asks to find the **instantaneous rate of change** of the function \( g(t) = 4 - t^2 \) at \( t = 3 \).

### Solution:

The instantaneous rate of change of a function at a given point is found using the **derivative** of the function.

1. First, we find the derivative of \( g(t) \).
   \[
   g(t) = 4 - t^2
   \]
   The derivative of \( g(t) \) with respect to \( t \) is:
   \[
   g'(t) = \frac{d}{dt}[4 - t^2] = 0 - 2t = -2t
   \]

2. Now, we evaluate the derivative at \( t = 3 \):
   \[
   g'(3) = -2(3) = -6
   \]

Thus, the **instantaneous rate of change** of \( g(t) \) at \( t = 3 \) is \( \boxed{-6} \).

Would you like a further explanation or details?

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Here are 5 related questions to expand your understanding:

1. What is the geometric meaning of the instantaneous rate of change of a function at a point?
2. How does the concept of the derivative relate to the slope of a tangent line at a point?
3. What are the rules for differentiating polynomials like \( g(t) = 4 - t^2 \)?
4. How would you calculate the average rate of change of \( g(t) = 4 - t^2 \) between \( t = 1 \) and \( t = 3 \)?
5. Can you find the second derivative of \( g(t) = 4 - t^2 \)? What does the second derivative tell you?

**Tip:** When solving for the instantaneous rate of change, always start by finding the derivative of the function before substituting the specific value of \( t \).

What is the instantaneous rate of change for g(t) = 4 - t^2 at t = 3?

Learn how to calculate the instantaneous rate of change for the function g(t) = 4 - t^2 at t = 3 using calculus and derivatives. Step-by-step explanation included. Ideal for high school or college students learning basic calculus concepts.

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